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mostsmooth
10-01-2003, 08:43 PM
the chance of rolling a total of 6 using two dice is 5/36,right? how do you figure the chance of rolling a 6 after x rolls?

mostsmooth
10-01-2003, 09:29 PM
[ QUOTE ]
the chance of rolling a total of 6 using two dice is 5/36,right? how do you figure the chance of rolling a 6 after x rolls?

[/ QUOTE ]
maybe i need to clarify better, what are the chances youd roll at least one 6 after 5 rolls?

Copernicus
10-02-2003, 12:58 AM
1-(1-31/36)^5

That is 1 minus the probability that you rolled no sixes in 5 rolls.

BruceZ
10-02-2003, 02:57 AM
what are the chances youd roll at least one 6 after 5 rolls?

1 - (31/36)^5 = 52.7%

That's 1 minus the probability of not getting a six 5 times in a row (C had 1 too many 1 minuses there /images/graemlins/blush.gif).

Now, if you roll until you get a 6, on which roll are you most likely to get it?

pudley4
10-02-2003, 03:18 PM
[ QUOTE ]

Now, if you roll until you get a 6, on which roll are you most likely to get it?

[/ QUOTE ]

The first /images/graemlins/smile.gif

Wake up CALL
10-02-2003, 03:42 PM
[ QUOTE ]
Now, if you roll until you get a 6, on which roll are you most likely to get it?

[/ QUOTE ]

The last roll of course.

Copernicus
10-02-2003, 05:07 PM
Or one too few "1-"s /images/graemlins/smile.gif Thanks...not sleeping enough lately trying to work and play tourneys at night!

Bozeman
10-02-2003, 06:38 PM
1st?

mostsmooth
10-02-2003, 06:43 PM
so the formula requires use the chances of a 6 not coming up ( 31/36) when calculating? theres no formula that uses the chance that it will come up? (5/36)?

Bozeman
10-02-2003, 07:23 PM
True. You can approximate the answer as n*p (n=#of tries, p= prob. on one try) if n and p are small. Basically, this multiply counts all the times where it happens more than once.

A conceptually different way to arrive at the accurate answer is:
P= p |chance of getting it on the first try
+(1-p)*p |chance of getting it on the second try and not getting it on the first
+(1-p)^2*p |chance of getting it on the third try and not 1st or 2nd
+(1-p)^3*p |etc.
+(1-p)^4*p
=1-(1-p)^5 |as given above with a little algebra
=52.6% for p=5/36

BruceZ
10-03-2003, 01:48 AM
Correct, pudley4 and Bozeman. The 1st roll is most likely to produce the first 6, with probability 5/36. All of the other rolls have lower probability because you have to NOT get a 6 on all the rolls before it AND then get a 6. That's harder. The probability for getting the first 6 on the Nth roll is (5/36)*(31/36)^(N-1) &lt;= 5/36, with = for N=1.

Now, next question. What is the average number of rolls necessary to throw a 6? There are 2 ways to solve this problem that I know of. One requires a little knowledge of calculus. The other requires only logic.

crockpot
10-03-2003, 05:42 AM
36/5.

Wake up CALL
10-03-2003, 11:19 AM
[ QUOTE ]
Correct, pudley4 and Bozeman. The 1st roll is most likely to produce the first 6, with probability 5/36. All of the other rolls have lower probability because you have to NOT get a 6 on all the rolls before it AND then get a 6. That's harder. The probability for getting the first 6 on the Nth roll is (5/36)*(31/36)^(N-1) &lt;= 5/36, with = for N=1.

Now, next question. What is the average number of rolls necessary to throw a 6? There are 2 ways to solve this problem that I know of. One requires a little knowledge of calculus. The other requires only logic.

[/ QUOTE ]

Sorry BruceZ, I know disagreeing with you can be hazardous to one's health but if you are going to ask this exact question "Now, if you roll until you get a 6, on which roll are you most likely to get it? Isn't the correct answer the last roll since you will then stop rolling?

Copernicus
10-03-2003, 12:34 PM
36/5 is correct, but what is the calculus approach? As a discrete function the only approach I know is to sum the infinite series.

I also agree that "last" is a better answer the way the question was originally phrased. While the probability of rolling the first 6 is highest on the first roll, you roll it on the last roll 100% of the time, including when the first roll is the last roll. /images/graemlins/crazy.gif

BruceZ
10-03-2003, 03:11 PM
Summing that particular infinite series requires calculus if you don't already know how to sum it.

How about showing some work for either method?

BruceZ
10-03-2003, 03:13 PM
I'll give you the same answer my old abstract algebra professors once gave me after I had submitted some dumbass answer. He said "if all I had found was a trivial answer like that one, I certainly would have looked around to see if I could have found a better answer". In other words, taking the easy way out may work in real life, but it doesn't get you very far in mathematics.

Wake up CALL
10-03-2003, 03:24 PM
[ QUOTE ]
I'll give you the same answer my old abstract algebra professors once gave me after I had submitted some dumbass answer. He said "if all I had found was a trivial answer like that one, I certainly would have looked around to see if I could have found a better answer". In other words, taking the easy way out may work in real life, but it doesn't get you very far in mathematics.

[/ QUOTE ]

I don't disagree with your professor BruceZ but I imagine he worded his questions better.

squiffy
10-03-2003, 03:40 PM
Wait, the probability of rolling 6 is the same for every roll. So how can the probability be greater on one roll than another?

BruceZ
10-03-2003, 03:49 PM
Nothing wrong with the wording of the question in either case, other than the fact that they afforded both an insightful answer and an unthinking answer. Those who chose to answer it insightfully hopefully learned something. Those who took the easy way out, as you did, only cheated themselves.

The roll you get the 6 is obviously the last roll. That was stated in the problem when we said that we roll until we get the 6. Your answer amounted to telling me what was already stated. Here's your F, now go stand in the corner. /images/graemlins/grin.gif

Nottom
10-03-2003, 04:00 PM
Because on the second roll and beyond, you have to have already not rolled a 6.

BruceZ
10-03-2003, 04:00 PM
See my solution above.

1st roll: probability = 5/36 = .139
second roll: probability = 31/36 * 5/36 = .120
third roll: probability = 31/36 * 31/36 * 5/36 = .103
...

The probabilities get smaller because you have to get something other than a 6 on all the previous rolls before you get a 6, and you still have to get the 6 on the partcular roll. On the first roll, you just have to roll the 6. The probability of rolling a 6 is the same on every roll, but the probability of getting the 1st 6 is not the same.

Good question, gold star for you (best I could do was a diamond).

Copernicus
10-03-2003, 04:10 PM
The work (a little difficult to write out here): The expected number of rolls is

(1*5/36 + 2*(31/36)*5/36 + 3*(31/36)^2*5/36......) = (factoring out the constant 5/36)

5/36* (1+ 2*31/36 + +3*(31/36)^2.....)=5/36*S, where S stands for the big sum.

For explanation, further simplify to S=(1+2r+3r^2....) where r = 31/36.

Then you play a little trick. r*S=(r+2r^2+3r^3...), ie every power of r has a coeffiecient 1 less than it does in S, (except there is no r^0 term)

therefore S-rS=(1+r^2+r^3....). The infinite series on the right is equal to 1/(1-r)=1/(1-31/36)=1/(5/36)=36/5.

ie S-rS=(1-31/36)*S =5/36*S = (see above) 36/5

Solving for S, S=(36/5)^2.

Way back at the beginning we factored out a 5/36 before the Sum, and the expected value was 5/36*S, so our EV is

5/36*(36/5)^2 = 36/5.

BruceZ
10-03-2003, 04:41 PM
And you didn't even really use calculus, just algebra (though I consider the infinite series itself to be part of calculus). You essentially found that the sum of 1 + r + 2r^2 + 3r^3 + ... to be 1/(1-r)^2 which is correct, but you got this just from the geometric series 1 + r + r^2 + ... = 1/(1-r) which more people are familiar with, and which you derive the same way. My method for doing the first sum involves taking a derviative of the geometric series to get the other series, hence the calculus.

Still waiting for an explanation of the logical solution from someone.

Copernicus
10-03-2003, 05:58 PM
Eureka..now I see the calculus approach...grazi!

well
10-04-2003, 08:22 AM
The algebraic approach seemed a little too obvious to look at. But the suggestion of coming up with an answer that uses logic started me to think.

Let's say we throw two dices M times, where M is a really large number. The amount of sixes expected would then be M*5/26. So we rolled until a six came up, and we did that about M*5/36 times (ignoring the last rolls, if the Mth outcome does not equal 6). And now, the average 'length' of a try would be about M/(M*5/36), which is 36/5.

Next Time.

BruceZ
10-04-2003, 01:23 PM
Or the way I look at it, which is equivalent to yours, suppose a very large number of people performed this experiment once, rolling until a six comes up. Then you know that out of all of those throws, a six must come up 5/36 of the time. But each person rolls exactly 1 six. So on average that the 1 six represents 5/36 of the total throws. Since 1 is 5/36 of 36/5, it took 36/5 throws on average to get a six.

Here's a variation of this problem that's a little easier to see. This is due to David. Suppose a billion chinese couples have children until they get a boy, and then they stop. How many children will be born on average? Assume each baby has an equal chance of being a boy or a girl. Since each couple will have exactly 1 child, there will be a billion boys born. But since the number of boys equals the number of girls, there will also be a billion girls, so there will be 2 billion children in all.

well
10-04-2003, 01:32 PM
It would be a little easier to see if you would have written
'...each couple will have exactly 1 boy' instead of '...one child'...

Next time.

BruceZ
10-04-2003, 01:59 PM
Since each couple will have exactly 1 <font color="red">child</font>, there will be a billion boys born.

Should be <font color="blue">boy</font>, right. Let's try this again:

Here's a variation of this problem that's a little easier to see. This is due to David. Suppose a billion chinese couples have children until they each get a boy, and then they stop. How many children will be born on average? Assume each baby has an equal chance of being a boy or a girl. Since each couple will have exactly 1 <font color="blue">boy</font>, there will be a billion boys born. But since the number of boys equals the number of girls, there will also be a billion girls, so there will be 2 billion children in all.

Copernicus
10-04-2003, 10:44 PM
"But since the number of boys equals the number of girls"

This is poorly worded as well. It is using the result (that there are a billion boys and a billion girls) to come to the conclusion that there are 2 billion total, which is a tautology.

The real issue of the problem is how does it happen that there ARE equal numbers of boys and girls. At first glance it would seem that there must be more Girls than Boys, since the sequences of children must be

B
GB
GGB
GGGB
GGGGB and so on, and it would seem that Girls overwhelm boys in these sequences. It is only when you get into the math of the infinite sequences that you can come to the conclusion that there are equal numbers of boys and girls and therefore 2 billion total.

BruceZ
10-05-2003, 12:27 PM
This is poorly worded as well. It is using the result (that there are a billion boys and a billion girls) to come to the conclusion that there are 2 billion total, which is a tautology. The real issue of the problem is how does it happen that there ARE equal numbers of boys and girls....It is only when you get into the math of the infinite sequences that you can come to the conclusion that there are equal numbers of boys and girls and therefore 2 billion total.

Absolutely not. The fact that there are an equal number of boys and girls on average in a large sample is a result of the fact that each child has an equal probability of being a boy or a girl as stated in the problem. This problem can be solved logically without any knowledge of the result of the series. That's the whole point of the problem. You need to think about this some more.

Here is the original thread (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=334 4&amp;Forum=probability&amp;Words=chinese&amp;Match=Entire%20P hrase&amp;Searchpage=0&amp;Limit=25&amp;Old=allposts&amp;Main=3344 &amp;Search=true#Post3344) by David Sklansky.